Solving the Arithmetic and Divisibility Problem

The question asks us to perform two mathematical operations and then determine which of the given options the final result is divisible by. First, we need to add 9435 to 7593. Then, we need to subtract 2607 from the sum obtained.

Step 1: Perform the Addition

We add the two numbers:

Let's perform the addition:

So, the sum of 9435 and 7593 is 17028.

Step 2: Perform the Subtraction

Next, we subtract 2607 from the sum, 17028:

Let's perform the subtraction:

The result after subtracting 2607 from 17028 is 14421.

Step 3: Check Divisibility of the Result

The final result is 14421. We now need to check if this number is divisible by any of the given options: 4, 10, 3, or 5. Let's examine the divisibility rules for each option:

• Divisibility by 4: A number is divisible by 4 if the number formed by its last two digits is divisible by 4. The last two digits of 14421 are 21. Is 21 divisible by 4? \(21 \div 4 = 5\) with a remainder of 1. So, 21 is not divisible by 4. Therefore, 14421 is not divisible by 4.
• Divisibility by 10: A number is divisible by 10 if its last digit is 0. The last digit of 14421 is 1. Since it is not 0, 14421 is not divisible by 10.
• Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3. The digits of 14421 are 1, 4, 4, 2, and 1. The sum of the digits is \(1 + 4 + 4 + 2 + 1 = 12\). Is 12 divisible by 3? Yes, \(12 \div 3 = 4\). Since the sum of the digits (12) is divisible by 3, the number 14421 is divisible by 3.
• Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5. The last digit of 14421 is 1. Since it is neither 0 nor 5, 14421 is not divisible by 5.

Conclusion

Based on the divisibility checks, the number 14421 is divisible by 3. Therefore, the result of the arithmetic operations is divisible by 3.

Revision Table: Arithmetic Problem Steps

Additional Information: Understanding Divisibility Rules

Divisibility rules are shortcuts to determine if a number is divisible by another number without performing the actual division. They are very useful in solving problems involving factors and multiples.

• Divisibility by 3: Sum of digits must be divisible by 3. Example: For 123, \(1+2+3=6\). Since 6 is divisible by 3, 123 is divisible by 3.
• Divisibility by 4: The number formed by the last two digits must be divisible by 4. Example: For 1516, the last two digits form 16. Since 16 is divisible by 4, 1516 is divisible by 4.
• Divisibility by 5: The last digit must be 0 or 5. Example: 175 ends in 5 (divisible by 5); 280 ends in 0 (divisible by 5).
• Divisibility by 10: The last digit must be 0. Example: 350 ends in 0 (divisible by 10); 990 ends in 0 (divisible by 10).